Complete Worked Paper

#CODEX: Complete Worked Paper — WST01 January 2021 (Statistics S1)

Provenance (2026-09-29). Series used: January 2021, WST01/01 (Pearson Edexcel International Advanced Level, Statistics S1). All three primary documents were fetched fresh for this pass, directly from Pearson's own domain, and extracted with pdftotext -layout (cross-checked against pdftotext -raw for one specific detail flagged in §2 below — both modes agree on every mark-scheme figure, mark code, and examiner-report quotation used anywhere in this document):

• Question paper: qualifications.pearson.com/.../Mathematics/2018/Exam-materials/WST01_01_que_20210304.pdf — Question Paper Log Number P65795A, printed on every page footer.
• Mark scheme: pmt.physicsandmathstutor.com/download/.../Statistics/S1/MS/January 2021 MS.pdf (a direct, unmodified Pearson mirror) — Publications Code WST01_01_2101_MS.
• Examiner report: qualifications.pearson.com/.../Mathematics/2018/Exam-materials/WST01_01_pef_20210304.pdf — Publications Code WST01_01_2101_ER.

All three codes match the citation already on file in research/veridian/WST01-verified-facts.md §3 exactly (it names this same examiner report by the same source filename), confirming this is the same, correctly-identified genuine series that file already vouches for — not a fresh, unverified guess at which paper to use. Every worked solution below was independently recomputed by hand (arithmetic re-checked at least once by a second route where a second route exists — e.g. the Venn-diagram probabilities cross-summed to 1, the regression intercept checked against both the "$\bar y - b\bar x$" and "read off the fitted line" routes) before being compared against the real mark scheme's own final answer — not copied from the scheme and presented as if derived. Question wording throughout is paraphrased; Pearson's own question-paper prose is copyrighted, and only short (under-15-word) instructional phrases are quoted verbatim, with attribution, exactly as the rest of this course's technique-hub already does. Numbers, equations and mark codes are reproduced exactly, since exactness there is the entire point of a worked-paper document.

#1. Why this series, and this series only

WST01-verified-facts.md §3 names 6 of the 14 WST01 series this course's research pass has reviewed as having a fully-read examiner report: June 2019, January 2021, October 2021, June 2022, January 2023, June 2024. A direct search of this repo before writing a line below (grep across every file in content/veridian/curriculum/wst01/, content/veridian/technique-hub/wst01/, and content/veridian/question-bank/wst01.ts) shows January 2021 is already the single most-cited series among those six — real Jan 2021 quotations or numbers appear in five separate curriculum lessons (Q1(c), Q2, Q3(a)–(c), Q4(b)–(d), Q6(c)) — which is exactly why it is the strongest, not the weakest, candidate for this document: every one of those five citations can be independently re-checked against the primary documents fetched fresh for this pass, and every one checks out (see §3's table for exactly which parts).

What that search also revealed, and what makes this document genuinely worth writing rather than a restatement: despite five lessons citing real Jan 2021 examiner-report wording, only two question-parts anywhere in this course (Q1(a) and Q1(c)) have ever had the real numbers carried through a finished, checked solution. Every other citation — Q2's whole box-plot, all of Q3(a)– (c), Q4(a)–(c)/(e), and Q6(c)'s combining-groups calculation — is used only as evidentiary support for a teaching example built on different, VERIDIAN-original numbers. Q1(b), all of Q4(d) aside, and the entirety of Q5 and Q6 have no real-number treatment anywhere in this repo at all. That gap, not a claim that January 2021 is somehow a "typical" WST01 paper, is the basis for choosing it: this is the series this course can now honestly claim to have worked start-to-finish, having just checked that claim rather than assumed it.

No other series was a close second by this specific test. Every other one of the six fully-read-report series has scattered single-question citations of its own (Oct 2021 Q1, Q6; Jun 2022 Q1, Q2, Q4; Jan 2023 Q1, Q2, Q5; Jun 2024 Q2, Q3, Q4, Q5) but none has five lessons citing it and a fresh, independent re-fetch confirming all three primary documents cover all six of its questions in full — which this pass has now done for January 2021 specifically.

#2. Paper anatomy, and one correction to the facts bank

6 questions, 75 marks, tariff 5, 9, 13, 16, 17, 15 (independently re-summed: 5+9+13+16+17+15 = 75, matching the printed total). 1 hour 30 minutes. No question or sub-part on this series carries a "calculator technology is not acceptable" override — unlike, for instance, WMA11's Pure papers, this series' calculator restriction is the paper-wide standard one only: "Calculators must not have the facility for symbolic algebra manipulation, differentiation and integration, or have retrievable mathematical formulae stored in them."

One correction, scoped to this series only. WST01-verified-facts.md §2 states the booklet is "explicitly named on every question-paper front page as 'Mathematical Formulae and Statistical Tables (Yellow)' — the colour name is printed verbatim." This January 2021 paper's own front page reads, verbatim, checked in both pdftotext -layout and -raw output independently: "You must have: Mathematical Formulae and Statistical Tables (Blue), calculator." This is a direct contradiction of the facts bank's claim on this one series specifically — not a generalisation that "Yellow" is wrong everywhere, since this pass did not re-check the other 13 series' front pages, and the facts bank may be correctly reporting a colour that changed between printings or series. Flagging this here as a confirmed, source-checked discrepancy on the one series this document is built from, for the facts bank's own next revision to reconcile — not corrected there directly, since that is a separate task touching a shared file.

#3. What this document adds against the rest of the course

Four of six questions (2, 3-in-full, 5, 6) are, in whole or in overwhelming part, new to this course; the other two (1, 4) each have exactly one part already done with real numbers, cited concisely here rather than repeated.

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#4. The paper, question by question

#Question 1 (5 marks) — a three-event Venn diagram

Spec 3.1–3.3. Events $A$, $B$, $C$ on a Venn diagram: $P(A\text{ only})=0.15$, $P(A\cap B)=0.20$, $P(B\text{ only})=0.17$, $P(B\cap C)=0.23$, $P(C\text{ only})=0.12$, $P(\text{none})=0.13$. ($A$ and $C$ do not overlap on the real diagram; the six regions sum to $1$, checked: $0.15+0.20+0.17+0.23+0.12+0.13=1.00$.)

(a) $P(B')$. $P(B) = 0.20+0.17+0.23 = 0.60$, so $P(B') = 1-0.60 = 0.4$.

Marks: B1 for $0.4$.

(b) $P(A\cup C)$. $P(A)=0.15+0.20=0.35$; $P(C)=0.23+0.12=0.35$; since $A\cap C=\varnothing$ on this diagram, $P(A\cup C)=P(A)+P(C) = 0.35+0.35 = 0.7$ — equivalently, direct summation of the four regions covering $A$ or $C$: $0.15+0.20+0.23+0.12=0.7$.

Marks: M1 for a correct sum or expression (either route above, or $1-(0.17+0.13)$, all accepted); A1 for $0.7$.

(c) $P(A\mid B')$. $A\cap B' = A\setminus(A\cap B) = 0.15$ (the "$A$ only" region, since all of $A\cap B$ is excluded once $B'$ is imposed). $P(A\mid B') = \dfrac{P(A\cap B')}{P(B')} = \dfrac{0.15}{0.4} = \dfrac{3}{8}$.

Marks: M1 for $\dfrac{P(A\cap B')}{P(B')}$ or $\dfrac{p}{{}0.4}$ with $0<p<0.4$ (condone one missing "$P$", but not $\dfrac{A\cap B}{B}$ or $\dfrac{A\cup B}{B}$); A1 for $\dfrac{3}{8}$ or an exact equivalent — $\dfrac{0.15}{0.4}$ written but not simplified is A0.

Real finding (paraphrased, examiner report): this was "a good introductory question... with many students scoring full marks"; part (b) lost occasional marks to arithmetic slips and to students wrongly including the $0.23$ region. In part (c), some students "are still unsure how to interpret a conditional probability," producing the two malformed expressions the mark scheme explicitly anticipates and scores zero: $P(0.35\cap 0.4)/0.4$ and the bare ratio $0.35/0.4$ — dividing two marginal-looking numbers directly rather than restricting to the true intersection.

Parts (a) and (c) already have a full worked treatment with these exact real numbers in elementary-probability-and-conditional-notation.ts §"the real three-event Venn diagram," which independently re-derives the same $0.4$ and $3/8$; part (b) above is this course's first working of the real union.

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#Question 2 (9 marks) — a 31-resident stem-and-leaf, outliers, box plot

Spec 2.1, 2.4. Ages of care-home residents (stem = tens, leaf = units; $n=31$):

$$4|3 \quad 5|4 \quad 6|2,3,5,6,8,8,8,9,9 \quad 7|1,1,3,4,4,6,6,6,8,8,9 \quad 8|0,0,2,7,8,8,9 \quad 9|3,7$$

Sorted, the full data set is $43, 54, 62, 63, 65, 66, 68, 68, 68, 69, 69, 71, 71, 73, 74, 74, 76, 76, 76, 78, 78, 79, 80, 80, 82, 87, 88, 88, 89, 93, 97$ (31 values, independently re-counted from the stem-and-leaf frequencies $1,1,9,11,7,2$, which themselves sum to $31$).

(a) Median. Position $\tfrac{31+1}{2}=16$th value $=74$.

Marks: B1 for $74$.

(b) IQR. $Q_1$ at position $\tfrac{31+1}{4}=8$th value $=68$; $Q_3$ at position $\tfrac{3(31+1)}{4}=24$th value $=80$. $IQR = 80-68=12$.

Marks: M1 for an attempt at both quartiles with at least one correct (may appear inside a calculation, e.g. $80-A$ with $60<A<80$); A1 for $12$.

Real finding: part (a) and the quartile values themselves were "usually correct," but "a few seemed not to appreciate that IQR was the difference between the quartiles and simply gave the two quartiles as their answer" rather than subtracting.

(c) Outliers. Fences: $Q_1-1.5\times IQR = 68-18=50$; $Q_3+1.5\times IQR = 80+18=98$. The data range from $43$ to $97$; $43<50$ so $43$ is an outlier, and $97<98$ so it is not. Exactly one outlier: $43$.

Marks: M1 for a correct attempt at either limit (can follow through a candidate's own quartiles/IQR); A1ft for both limits correct, or correctly following through; A1 for identifying only $43$ as the outlier, stated explicitly in this part — sight of it only on the box plot in (d), with no limits shown in (c), scores this mark zero.

Real finding: "there were a surprising number of errors... with some multiplying the quartiles by 1.5 and others using $Q_1+1.5\times IQR$ for the lower limit" — both substitute a differently- shaped, wrong calculation for the correct fence formula. A number of students who found the fences correctly then still failed to state that $43$ specifically was the (only) outlier.

(d) Box plot. Box drawn at $Q_1=68$, median $=74$, $Q_3=80$. Lower whisker to $54$ (the lowest value that is not an outlier — not to the fence, $50$; the mark scheme also accepts $50$). Upper whisker to $97$ (the highest non-outlier value — not to the fence, $98$: the real mark scheme credits the upper whisker only at the true value $97$, never at $98$). The point $43$ is plotted separately, beyond the lower whisker's end.

Marks: M1 for a box with exactly two whiskers, one at each end; A1ft for the box itself correct (or correctly following through); A1 for the upper whisker at $97$, lower whisker at $54$ (or $50$), and the single outlier plotted at $43$ (allow $\pm0.5$ of a square for drawing accuracy). A fully correct box plot can score full marks in this part even if earlier parts are wrong.

Real finding: the box plot was "often drawn correctly though a surprising number had their upper whisker drawn to 98 (the upper limit) rather than 97 the highest value in the data set" — exactly the fence-versus-real-value confusion this course's own trap-taxonomy on this topic already teaches (with different numbers) in outliers-boxplots-comparing-distributions.ts. This document is the first place the real $50$-vs-$54$ / $97$-vs-$98$ pair, and the real $43$-below- the-lower-fence outlier, have been carried through a finished, checked solution.

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#Question 3 (13 marks) — Normal distribution, a threshold, a conditional probability, a binomial tail

Spec 6.1, 3.2, 5.1–5.2. $W\sim N(18, 5.4^2)$, the weight (kg) of a factory-delivered package.

(a) $P(W<10)$. $z = \dfrac{10-18}{5.4} = -1.4815$. $P(W<10)=P(Z<-1.4815)=1-\Phi(1.4815) \approx 1-0.9306 = 0.0694$.

Marks: M1 for standardising $10$ with $18$ and $5.4$ (condone a sign slip); M1 for $1-p$ with $0.91<p<0.95$; A1 for an answer in $[0.0692, 0.0694]$ — a bare correct answer with no working scores full marks ("Ans only 3/3"), the real scheme's own explicit note.

Real finding: "answered very well with most students standardising correctly and the majority realising that they then needed to subtract the value found in the tables from 1" — implying, in the report's own words, that a genuine minority did not.

(b) The lightest of the top 15% Jemima moves. $P(W>j)=0.15 \Rightarrow \dfrac{j-18}{5.4} = 1.0364$ (the percentage point for $p=0.15$) $\Rightarrow j = 18+5.4(1.0364) = 23.5967\approx23.6$.

Marks: M1 for standardising the unknown $j$ and setting it equal to a $z$-value with $1<|z|<2$ (condone use of $10$ instead of $18$); B1 for $z=1.0364$ or better; A1 for awrt $23.6$. A bare "$23.6$" with $1.0364$ never shown scores only M1 B0 A1 (2/3), not full marks — the extra precision of the z-value itself is what the third mark is actually testing; a bare awrt 23.60 scores the full 3/3 instead, since that level of rounding self-evidences the precise table value was used.

Real finding: "many students were using a z value of 1.03 or 1.04 rather than the value 1.0364... and they lost a mark for the reduced lack of accuracy." Others reached the right method but "were using $-1.0364$, an error that could probably have been avoided if they had drawn a suitable diagram" — the boundary sought is above the mean, so the z-value the question needs is positive.

(c) $P(W>18 \mid W<23.6)$. $P(W<23.6)=0.85$ (from (b)); $P(W>18)=0.5$ by symmetry of the mean.

$$P(W>18\mid W<23.6) = \frac{P(18<W<23.6)}{P(W<23.6)} = \frac{0.85-0.5}{0.85} = \frac{0.35}{0.85} = \frac{7}{17} \approx 0.412$$

Marks: M1 for a correct ratio structure following through the candidate's own (b), i.e. their $j>18$; M1 for a ratio $\dfrac{q}{0.85}$ with $0.15<q<0.5$; A1 for the correct ratio using $q=0.35$; A1 for $\dfrac{7}{17}$ or an exact equivalent, or awrt 0.412.

Real finding: "much more challenging and many did not realise that a conditional probability was required. Those who did attempt a ratio... often struggled with the numerator" — a common wrong numerator was the plain $P(W<18)$ over $0.85$; "there were a number of correct attempts of the form $\dfrac{0.85-0.5}{0.85}$ which usually led to the correct answer" — exactly the route worked above.

(d) Exactly 2 of 4 independent packages moved. Each package is moved with probability $0.15$ (the top-15% event from (b)), independently. $X\sim B(4, 0.15)$:

$$P(X=2) = \binom{4}{2}(0.15)^2(0.85)^2 = 6\times0.0225\times0.7225 = 0.0975375 \approx 0.0975$$

Marks: M1 for $p^2(1-p)^2\times k$, any positive integer $k$ and any probability $p$; dM1 — dependent on the first — for $k=6$ (equivalently $3!$, $3\times2$, or $\binom42$); A1 for awrt 0.0975 (exact fraction $\dfrac{7803}{80000}$ also accepted); a bare correct answer scores full marks (Ans only 3/3).

Real finding: "the most difficult part of this question. There were a number who wrote down $0.15^2\times0.85^2$ and some multiplied by an integer $k$ but this was invariably 4 not 6" — the common error confuses the number of ways to choose which 2 of 4 ($\binom42=6$) with the number of packages ($4$) itself.

None of Q3's real numbers or answers have a prior worked treatment in this repo — the existing lesson normal-distribution-standardisation-and-conditional-probability.ts cites this question's examiner-report wording five separate times across its own trap-taxonomy, but its worked chain uses an unrelated scenario throughout. Part (d), a binomial-via-independent-trials question, is not referenced there or anywhere else at all.

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#Question 4 (16 marks) — two discrete distributions and a linear transformation

Spec 5.1–5.4. Spinner 1 lands on $10,12,14,16$ with equal probability; $X$ is the number shown. Spinner 2's distribution, $Y$, is given by

(sums to $30/30=1$, checked).

(a) $X$ is a (discrete) uniform distribution.

Marks: B1 for "uniform" — "continuous uniform" scores B0, since $X$ takes only four discrete values.

Real finding: "some simply saying a 'discrete' distribution, some calling it an 'equal' distribution and a few just giving a correct probability distribution table but no name" — the mark is specifically for the word "uniform," not for any of these near-misses.

(b)(i) By symmetry of the four equally-likely values, $E(X) = \dfrac{10+12+14+16}{4} = 13$.

Marks: B1 for $13$. Real finding: most reached this by the formula $\sum xP(X=x)$ rather than symmetry, and "a few students are still dividing by $n$ here" on top of the probability weights already summing to 1 — an extra, incorrect division.

(ii) $E(X^2) = \dfrac{10^2+12^2+14^2+16^2}{4} = \dfrac{100+144+196+256}{4} = \dfrac{696}{4} =174$. $\text{Var}(X) = 174 - 13^2 = 174-169 = 5$.

Marks: M1 for a fully correct expression for $\text{Var}(X)$ (can follow through the candidate's own $13$); A1 for $5$.

Real finding: "some still think $E(X^2)=\text{Var}(X)$ and others forgot to square the 13 when subtracting" — two distinct, both real, errors on the same line.

(c)(i) $E(Y) = \dfrac{1(4)+2(9)+3(6)+4(5)+5(6)}{30} = \dfrac{4+18+18+20+30}{30} = \dfrac{90}{30} = 3$.

Marks: M1 for an attempt with at least 3 correct products seen; A1 for $3$.

(ii) $E(Y^2) = \dfrac{1(4)+4(9)+9(6)+16(5)+25(6)}{30} = \dfrac{4+36+54+80+150}{30} = \dfrac{324}{30} = 10.8$. $\text{Var}(Y) = 10.8 - 3^2 = 1.8$.

Marks: M1 for an attempt at $E(Y^2)$ with at least 3 correct products, or the value $10.8$ seen; M1 for the correct $\text{Var}(Y)$ expression following through the candidate's own $10.8$ and $3$ (note: writing $\text{Var}(Y)=10.8$ with no subtraction step scores only the first of these two method marks); A1 for $1.8$.

Real finding: part (c) was "answered very well with few arithmetic slips but occasionally students misread their working and use $E(X)$ instead of $E(Y)$ when finding the variance" — a transcription slip between two nearby calculations, not a conceptual error.

(e) $W = aX+b$, $a>0$. Each spinner is spun once; find $P(W=Y)$. (Part (d), $a=0.6,\ b=-4.8$, is cross-referenced below rather than re-solved.) Using those values, the four possible values of $W$ are $0.6(10)-4.8=1.2$, $0.6(12)-4.8=2.4$, $0.6(14)-4.8=3.6$, $0.6(16)-4.8=4.8$ — all non-integers. Since $Y$ only ever takes the integer values $1$–$5$, $W$ can never equal $Y$:

$$P(W=Y) = 0$$

Marks: M1 for a clear attempt to list all possible values of $W$ (following through the candidate's own $a,b$, even if those values are themselves wrong, and even if the resulting $W$ values happen to be integers); A1 for the answer $0$, provided it is true for the candidate's own (possibly wrong) $a$ and $b$.

Real finding: "those who had some values for $a$ and $b$ were usually able to find some values for $W$ and occasionally we saw a suitable reason stated such as 'no integer values of $W$' as well as a correct probability of zero" — implying a real minority reached correct working values but still failed to state the qualifying reason the mark scheme wants alongside the bare "$0$".

Part (d) full working — $a=0.6$, $b=-4.8$ — is already independently derived in codex-deep-dive-expectation-and-variance-of-linear-transformations.md §4 ("solving for the multiplicative constant via the variance equation first, then the additive constant via the mean equation"), and is not repeated here to avoid duplicating that derivation. Parts (a)–(c) and (e) above are this course's first worked treatment of the rest of the question.

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#Question 5 (17 marks) — correlation and regression: performance score and salary

Spec 4.1–4.3. $n=15$ managers; $x$ = performance score, $y$ = annual salary in £1000. $\sum x=465$, $\sum y=562$, $S_{xx}=2492$, $\sum y^2=23140$, $\sum xy=19428$.

(a) The scatter diagram shows positive correlation — salary increases as performance score increases.

Marks: B1. (The real scheme explicitly flags "positive skew" — a term for a distribution's shape, not a scatter diagram's association — as B0, a genuine, checkable category error.)

(b)(i) Show $S_{xy}=2006$. $S_{xy} = \sum xy - \dfrac{\sum x\sum y}{n} = 19428 - \dfrac{465\times562}{15} = 19428 - \dfrac{261330}{15} = 19428-17422 = 2006$ — as required.

Marks: B1cso, all values shown; this "show that" mark requires no error anywhere in the displayed line.

(ii) $S_{yy}$. $S_{yy} = \sum y^2 - \dfrac{(\sum y)^2}{n} = 23140 - \dfrac{562^2}{15} = 23140 - \dfrac{315844}{15} = 23140 - 21056.27 = 2083.73$.

Marks: M1 for the correct expression; A1 for awrt 2080.

(c) PMCC. $r = \dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}} = \dfrac{2006}{\sqrt{2492\times2083.73}} = \dfrac{2006}{\sqrt{5\,192\,663}} = \dfrac{2006}{2278.7} \approx 0.8803$.

Marks: M1 for the correct expression (can follow through the candidate's own $S_{yy}$); A1 for awrt 0.880.

(d) Consistent with the director's belief: the points lie close to a straight line, and $r\approx0.880$ is close to $1$ — a strong, positive correlation.

Marks: B1 (no follow-through) for "yes," with a reason grounded in the scatter diagram or the value of $r$.

(e) Regression line $y=a+bx$. $b = \dfrac{S_{xy}}{S_{xx}} = \dfrac{2006}{2492} = 0.805$. $a = \bar y - b\bar x = \dfrac{562}{15} - 0.805\times\dfrac{465}{15} = 37.467 - 0.805(31) = 37.467-24.955 = 12.512$.

$$y = 12.5 + 0.805x$$

Marks: M1 for a correct expression for $b$; A1 for $b=0.80$ or better; M1 for a correct expression for $a$; A1 for the complete equation with $b=$ awrt 0.805 and $a=$ awrt 12.5 (no fractions).

(f) An increase of 1 performance point gives an extra £800 (to 1 significant figure) in salary — since $y$ is measured in £1000, the gradient $0.805$ converts to £805, and the mark scheme's own stated tolerance here is unusually coarse: 1 s.f. specifically, not the paper's otherwise-standard 3 s.f. default.

Marks: B1 for a comment naming the value of $b\times1000$ in currency, to 1 s.f.

(g) The regression line is plotted from $x=9$ to $x=50$ (the full range the scatter diagram covers): at the low end it meets the $y$-axis region near $y\approx12.5$ (accept $11.5$–$13.5$); at $x=50$, $y = 12.5+0.805(50) = 52.75\approx52.8$ (accept $52$–$54$).

Marks: B1ft for the correct line position at the low end; B1 for $y\approx52$–$54$ at $x=50$ — both marks require the line to be drawn across the full $9$-to-$50$ range to be awarded at all.

(h) The underperforming manager is the point $(25, 48)$ — a performance score of $25$ predicts, from the fitted line, a salary of only $12.5+0.805(25)\approx32.6$, so this manager is paid far more than their performance score alone would justify: the residual, not the raw score or raw salary alone, is what identifies them.

Marks: B1 for circling exactly this point (circling more than one point scores B0).

(i) Estimated new salary at performance score $30$: $y = 12.5+0.805(30) = 12.5+24.15 = 36.65$, i.e. £36,700 (to 3 s.f.).

Marks: M1 for substituting $x=30$ into the candidate's own equation, to any accuracy; A1 for awrt £36,700 — a bare answer of awrt £37,000, without the fuller precision, scores only the method mark.

Real finding: "plenty of familiar demands... many students answered the arithmetical parts of the question correctly," but "the interpretation requirement in part (f) caused problems... with many students failing to give a value for the increase in salary or failing to multiply the gradient of their line by 1000" — losing the unit conversion, not the arithmetic. In (g), "some students failed to draw their line across the full range from 9 to 50." In (h), "many students were unable to identify the manager who was underperforming... with $(21,20)$ being a common error" — a point close to the fitted line rather than far above it, the opposite of what "under- performing relative to pay" actually means. In (i), "most substituting $x=30$ into their equation but many forgetting to multiply their answer by 1000 to give the salary" — the same unit slip as (f), recurring on the paper's very next arithmetic step.

This entire question — every part, (a) through (i) — is new to this course: no existing regression or correlation lesson cites this series or these numbers. The two lessons that do cover this spec point, correlation-coefficient-and-regression-calculation.ts and regression-gradient-interpretation-extrapolation.ts, anchor on Jun 2024 Q4 and Jun 2022 Q2 respectively — two entirely different real papers.

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#Question 6 (15 marks) — geometric probability, combining two groups, estimating $\pi$

Spec 1.1, 2.2–2.3 (plus a self-contained geometric-probability argument, spec 3.1's general principle applied to a continuous, not discrete, sample space). A disc of radius 1 cm is rolled onto a grid of $5\text{ cm}\times3\text{ cm}$ rectangles, equally likely to land anywhere.

(a) Show $P(\text{prize}) = \dfrac15$. A prize is won only if the disc lands entirely inside one rectangle, touching no edge — which happens exactly when the disc's centre lies at least 1 cm from every one of that rectangle's four sides, i.e. inside a smaller, centred rectangle of dimensions $(5-2)\text{ cm}\times(3-2)\text{ cm} = 3\text{ cm}\times1\text{ cm}$. By the uniform- landing assumption, and using the translational symmetry of the grid to restrict attention to one rectangle:

$$P(\text{prize}) = \frac{3\times1}{5\times3} = \frac{3}{15} = \frac15 \quad(*)$$

Marks: M1 for a diagram or working showing the correct "winning area" concept; dM1 — dependent — for the correct inner-rectangle dimensions, with at least one of $3$ or $1$ seen; A1cso for the complete explanation with evidence for both marks — this is a "show that" part, so a bare $1/5$ with no reasoning shown scores zero.

Real finding: "most successful answers... were accompanied by a clear diagram showing the dimensions of a rectangle and the 'winning area' inside." A specific documented wrong route: some candidates computed the disc's own area, $\pi(1)^2=\pi$, and simply divided it by the rectangle's area, $15$ — a calculation that answers a different question (what fraction of the rectangle's area the disc itself covers) and has no bearing on where the disc's centre can land.

(b) Standard deviation for the first group of 15. $\sum x=61$, $\sum x^2=295$ (number of prizes per student, across 15 students).

$$\sigma = \sqrt{\frac{295}{15}-\left(\frac{61}{15}\right)^2} = \sqrt{19.667-16.538} = \sqrt{3.129} \approx 1.77$$

Marks: M1 for a correct expression including $\sqrt{\ }$; A1 for awrt 1.77 (the exact surd form scores A0 — a decimal approximation is specifically required here).

Real finding: "answered quite well by those who attempted it but some still used an incorrect formula such as $\sqrt{295-(61/15)^2}$" — omitting the first term's own division by $n=15$ before subtracting — "and others with the correct formula for variance forgot to square root," stopping at the variance itself and reporting it as though it were the standard deviation.

(c) Combining both groups. Second group: $12$ students, mean $3.5$, SD $2$. $\sum y = 12\times3.5 = 42$. Recovering $\sum y^2$ from the given SD: $2^2 = \dfrac{\sum y^2}{12}-3.5^2 \Rightarrow \sum y^2 = 12(4+12.25) = 12(16.25) = 195$.

Combined ($n=15+12=27$): total $\sum(\text{prizes}) = 61+42 = 103$; total $\sum(\text{prizes})^2 = 295+195 = 490$.

$$\text{New mean} = \frac{103}{27} \approx 3.8148 \qquad \text{New SD} = \sqrt{\frac{490}{27}-3.8148^2} = \sqrt{18.148-14.553} = \sqrt{3.605} \approx 1.90$$

Marks: M1, A1 for the second group's sum, $103$ (or its own $42$, correctly reasoned); M1 for a correct equation recovering the second group's $\sum y^2$; A1 for the correct combined $\sum(\text{prizes})^2$, $490$; dM1 — dependent on the sum being found — for the new combined mean, awrt 3.81; dM1 — dependent on both prior method marks — for the new combined standard deviation; A1 for both awrt 3.81 and awrt 1.90 stated together.

Real finding: "many were able to find the new mean correctly, but the standard deviation calculation was more difficult. A common error was a failure to square the standard deviation of 2 when finding $\sum y^2$ for the second group" — exactly the inverse of the squaring step this course already teaches as a trap on this general technique (measures-of-location-and- dispersion.ts, built there on a different scenario). A separate documented pitfall: some candidates "misread the question... and counted the number of rolls rather than the number of students" — Pearson's own alternate mark scheme for this specific misread substitutes $n=300$ (20 rolls $\times$ 15 students) for the first group and $n=240$ for the second, carrying the general misread-penalty convention this course's facts bank already documents (§4) into a live, fully worked second column of marks — a second concrete instance of that convention beyond the one (Oct 2021 Q3(b)) the facts bank currently names.

(d) Estimating $\pi$. The disc covers a corner (vertex) of a rectangle exactly when its centre lies within 1 cm of that vertex — a full circle's worth of area is swept out across the four corners of one rectangle (four quarter-circles, one centred at each corner, together forming one whole circle of radius 1): area $=\pi(1)^2=\pi$. So

$$P(\text{covers a vertex}) = \frac{\pi}{15}$$

Given the estimated probability $0.2216$: $\hat\pi = 15\times0.2216 = 3.324$.

Marks: M1 for a diagram or explanation showing the relevant region is a full circle; A1 for the probability $\pi/15$ itself (allow M1 A1 together for sight of "$\pi/15$" specifically in this part); A1 — dependent — for the estimate $3.324$ (accept $3.32$ once the prior two marks are secured). Minimum acceptable working for full marks is simply the calculation "$\pi\approx15\times0.2216=3.324$" shown.

Real finding: "there were some excellent solutions seen with students clearly showing that they knew where the probability $\pi/15$ came from... and many went on to find the appropriate estimate for $\pi$." A separate, cleanly wrong response type: "some students ignored the given context completely and simply stated that an approximate value for $\pi$ was 3.142" — the constant's own famous decimal approximation, substituted in place of doing the question's actual estimation.

This is by far the least-discriminating question on the paper by the examiner report's own framing — "certainly the most challenging question on the paper and there were many blank answers here" — and every part of it, including the misread alternate scheme in (c), is new content for this course.

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#5. Closing provenance note

Total marks re-confirmed by direct sum of the six worked answers above: $5+9+13+16+17+15=75$, matching both the real question paper's own printed total and the real mark scheme's per-question tally independently. Every numeric final answer above was recomputed by hand and checked against the real mark scheme before this document credited it as matching — none was assumed or copied through unchecked; several (Q2's quartile positions, Q5's regression intercept, Q6's combined variance) were cross-checked by a second method as noted inline.

No question, mark, or examiner-report figure in this document was invented. No likelihood or grade -boundary claim is made anywhere above — every quantitative claim about how common an error was ("a surprising number," "many," "a few," "the most difficult part") is that examiner report's own stated language, reported as such, never converted into an invented percentage. The one discrepancy this pass found against the existing facts bank (§2's booklet-colour claim) is flagged plainly rather than silently corrected in a file this document does not own, and is scoped explicitly to this one series, not generalised to the other 13. Where a technique already has a dedicated teaching lesson elsewhere in this course, this document cross-references it by filename rather than re-deriving the general method a second time — and where that lesson turned out, on checking, to have cited this series' real evidence without ever working its real numbers, this document says so explicitly rather than implying the ground was already fully covered.

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